mathdojofandomcom-20200214-history
Julian Homework 1
Comments: 2.2: Worried you forgot about proving multiplication is closed, but you saved that for last. Spot on. 2.8: Concise and nice! 3.1i: Good. 3.1ii: Even better. 3.2: Dude, I graded Math 109 and SO MANY PEOPLE FORGOT TO CHECK INVERSE IS IN THE SET. Finally someone gets it right. 3.3: Cannot define an nth complex root since we are guarenteed n different values (ex. 4th complex root of 1 is -1, 1, i, -i). So this is not a function (and in math, we cannot argue with non-functions because of an issue called the number being non well defined). However, what you could have done is kept the exponent: Ex. for inverses: let x be such that x^n=1. Then let y = 1/x. y^n = (1/x)^n = 1/(x^n)=1/1=1 (essentially the same arguement you did, except I kept the exponent). 3.10: YES! YES! Prime example of where number theory comes in (no pun intended). P.S, you could have used Bezout's identity to prove the inverse exists : ) . 4.2: Dude. The identity in multiplication is not 0. Fail. 4.4: GOOD! Because if gx_1=gx_2, then multiplying both sides on the left by g's inverse, x_1=x_2. YES! 4.5: Could have used simpler, less rigorous counting arguement but ok. Alright, base case holds because 0 number of one element of order 2. For inductive step (by the way, | reads as divides. Not divisible by 2). I am kinda concerned about the rigorousness of building a new group by adding two elements. Can any group of even order be built by taking some group and then adding precisely two more elements such that it does not a super explosion of brand new elements. How do we know that G + {x,y} is still a group? Definitely a simpler arguement out there, maybe with proof by contradiction and fixing the values. 4.6: YES! 4.8: Noooooooo. So close. xy=yx is NOT always true and is true in an ABELIAN group (multiplication need not be commutative). However, nested in here, you did show that yx(y^-1) had the same order as x (by writing out and using associativity to regroup and cancel. For the xy having same order as yx, same trick with writing out and associativity. Homework 1 Supplement 5.5. Forward direction, trivial, duh. Converse (good, math vocab!), good you noticed that x can equal y. GREAT, you understand the key behind the word finite: powers of y cannot all be unique. Associativity is also obvious because associativity holods already in G. Great proof! (In last sentence, just because I like to nitpick and jetlag, The first part...Thus "because of H's status as a group..should be replaced with"..."Because if xy in H....inverse exist... H<=G. Remove H's status as a group part because you are concluding it is a group. 5.7: Forgot to argue the lynchpin, why x, x^2, ...x^m-1 have finite order. But this is simple by commutivity of exponent multiplication: (x^k)^n= x^(kn)=(x^n)^k= e^k= e. Everything else, excellent! 5.10. Good. Since you know FTA, gcd part easy to prove. 6.1. Good. Sorry to make you do this one, but every math major has to do this once in their life (permutation multiplication table). 6.3. Whoa, almost 100% of students got this wrong and wrote 3!. But for every permutation of 2,5,7 you have 6! permutation of 1,3,4,6,8,9. So it is 4320. Whoa, incredible job. As a small note, you forgot to prove why inverse exist (this is easy though, because if you are only permute 3 objects, to get back to the original state, you only need to permute amoung those 3 objects). 6.11. Good you used theorem, some people actually wrote out entire computation. grouptheoryhomework1(julians)-001.jpg|Julian's Homework 1 p.1 grouptheoryhomework1(julians)-002.jpg|Julian's Homework 1 p.2 grouptheoryhomework1(julians)-003.jpg|Julian's Homework 1 p.3 index.16859.2112312378168590.jpg|Julian's Homework 1 p.4 index.16859.2112312378168591.jpg|Julian's Homework 1 p.5